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Integer Illation From Divisors (Posted on 2016-01-27) Difficulty: 3 of 5
Find all possible values of a positive integer M such that:
M has precisely 16 positive integer divisors D(1), D(2), ......, D(16), and:
1 = D(1) < D(2) < ......< D(16) = M, and:
D(6) = 18, D(9) – D(8) = 17

No Solution Yet Submitted by K Sengupta    
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Solution Solution Comment 2 of 2 |
16 must be the product of one more than the exponents in the prime factorization of M.

So M must have one of the following forms.
(i) p^15
(ii) p^7*q
(iii) p^3*q^3
(iv) p^3*q*r
(v) p*q*r*s

Since D(6)=18=2*3^2 we can rule out (i) and (v).  Since 18 itself has 6 factors, 2^2 can't be a factor so this rules out (iii).

(ii) would require the number to be 2*3^7 whose first 9 factors are
1,2,3,6,9,18,27,54,81.  81-54=27 not the required 17 so (ii) is out.

So the number must be of the form (iv).  Specifically 2*3^2*r where r is a prime > 18.
We need this prime to slot into the above list either before or after the 54 and give the difference 17.

It turns out that both options work
54-37=17 and 71-54=17.

So M is either
2*3^3*37=1998
or
2*3^3*71=3834

Edited on January 27, 2016, 10:21 am
  Posted by Jer on 2016-01-27 10:18:51

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